Since the sum of digits are equivalent to the number in mod9, We easily the number is equivalent to 0or1 in mod9. (case 1)$n \equiv 1$ Let the number of digit be m. then obviously $n \leqq (9m-8)^2$ that implies m<4. by simple calculating we get n=1. (case 2)$n \equiv 0.$ Similar way to case 1. n=0,81. Then the solution is n=0,1,81.

Bigwood wrote: Since the sum of digits are equivalent to the number in mod9, We easily the number is equivalent to 0or1 in mod9. I don't think so....a square can be congruent with $ 0,1,4,7 $ modulo $ 9 $... and if their sum is divisible by $ 9 $ that doesn't mean that each number is divisible by $ 9 $. if $ n $ has $ t $ digits then $ 81t \geq n>10^{t-1} $ so $ 81t>10^{t-1} $. this is false for $ t \geq 4 $. so $ t=1,2,3 $. now is easy.

To anonymouslonely. Read again, it is said that the number equals the square of the sum of its digits, not the sum of the squares of its digits, as you seem to have read it. So since $n\equiv S(n) \pmod{9}$, from $n=S(n)^2$ follows $n\equiv n^2 \pmod{9}$, i.e. $n(n-1) \equiv 0 \pmod{9}$, therefore $n \equiv 0,1 \pmod{9}$, as claimed.

How can one put those problems from Lusophon MO in the contest section? I mean, there is no section for this Olympiad. Could someone tell me how to create a new section?

Only RManagers (such as myself) can do so. Post the links to the problems in the International Contests forum or the National Olympiads forum, and then we'll add them for you. EDIT: Is this a Portuguese contest?

bluecarneal wrote: Only RManagers (such as myself) can do so. Post the links to the problems in the International Contests forum or the National Olympiads forum, and then we'll add them for you. Thanks bluecarneal wrote: EDIT: Is this a Portuguese contest? Not at all. This is a contest for portuguese-speaking countries (Portugal, Brazil, Mozambique, etc.) This year the problems were very easy. In fact I'm ashamed of posting problems 1, 3 and 4 But I think it's better to post all the problems.

Ok. I assume the contest is over, otherwise you'd be cheating. Just post the links to the problems in the forum I mentioned, and I'll add them to the International Section.

again Gosk-OC inequality: $s(n)<9lgn$ is useful.

littletush wrote: again Gosk-OC inequality: $s(n)<9lgn$ is useful. Hey, I think it is $9log n$. But it is undoubtedly good idea. Thanks

Bigwood wrote: littletush wrote: again Gosk-OC inequality: $s(n)<9lgn$ is useful. Hey, I think it is $9log n$. But it is undoubtedly good idea. Thanks what do you mean $logn$?that's $log_{10}n$?

littletush wrote: Bigwood wrote: littletush wrote: again Gosk-OC inequality: $s(n)<9lgn$ is useful. Hey, I think it is $9log n$. But it is undoubtedly good idea. Thanks what do you mean $logn$?that's $log_{10}n$? Sorry, yes. If you thought it should be something other than $log_{10} n$, wud you mind defining "$lgn$?? Thanks.

I see.then $lgn$ is the same as $logn$.

Bigwood wrote: Since the sum of digits are equivalent to the number in mod9, We easily the number is equivalent to 0or1 in mod9. (case 1)$n \equiv 1$ Let the number of digit be m. then obviously $n \leqq (9m-8)^2$ that implies m<4. by simple calculating we get n=1. (case 2)$n \equiv 0.$ Similar way to case 1. n=0,81. Then the solution is n=0,1,81. 81 is not squaredigital..

We can easily check the cases where $n$ has $1, 2, 3$ or $4$ digits. And, by easily I mean using the fact that: - if $n$ has $m$ digits, $[s(n)]^2$ is, at most, equals to $(9m)^2$ For the cases where $n$ has more than $5$ digits, apply $log()$ in both sides and notice that: - In right hand side: $log(n)\ge m-1$ - In left hand side: $log(9m)^2\ge log[s(n)]^2$ Finally, use induction over $m$ to show that: $m-1\ge log(9m)^2$ By the way, $n=0, 1$ or $81$.

What a fun problem. Let $n=\overline{a_{k}a_{k-1}\dots a_{0}}$ so we want all the solutions for the equation $(a_{k} + a_{k-1} +\dots a_{0})^2=a_{k}10^k + a_{k-1}10^{k-1}+\dots a_{0}$. Now let us notice that $a_{k}10^k + a_{k-1}10^{k-1}+\dots a_{0}\geq 10^k$ and $(a_{k} + a_{k-1} +\dots a_{0})^2\le (9k)^2=81k^2$, and we can easily prove by induction that $10^k> 81k^2$ for all natural numbers $k>5$ which means we just need to find solution for $k=1,2,3,4$. Case $k=1$ is trivial because its easy to see that $0$ and $1$ are the only solutions. Case $k=2$ $(a_{0}+ a_{1})^2=10a_{1}+ a_{0}$, in fact all $squaredigital numbers$ are perfect squares so here we just need to check the list $16,25,36,49,64,81$ and the only sotuion is $81$. Case $k=3$ I could not find a smart way to prove that there are no solutions here but since $(a_{0}+ a_{1}+ a_{2})^2\le27^2=729$ then $a_{2}\le 7$ so $(a_{0}+ a_{1}+ a_{2})^2\le25^2=625$ so $a_{2}\le6$ and doing it again and checking some cases will have $a_{2}\le3$ but after that I just checked the rest of the numbers. Case $k=4$ $(a_{0}+ a_{1}+ a_{2}+ a_{3})^2\le 36^2=1296$ so $a_{3}=1$ but then again $(a_{0}+ a_{1}+ a_{2}+ a_{3})^2\le(1+27)^2<1000$ but since $n$ is a $4$ digit number this case does not have solutions. So the only $squaredigital numbers$ are numbers are $0,1$ and $81$. $\square$